Question

Assertion :If $a+b+c=0$ and $a,b,c$ are rational, then the roots of the equation $(b+c-a)x^2+(c+a-b)x+(a+b-c)=0$ are rational . Reason: Discriminant of $(b+c-a)x^2+(c+a-b)x+(a+b-c)=0$ is a perfect square .

• A. Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I .
• B. Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I .
• C. Statement-I is true, Statement-II is false .
• D. Statement-I is false, Statement-II is true .

Answer 1

The correct option is A Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I .

Given $a+b+c=0$ and $a,b,c$ are rational .
$(b+c-a)x^2+(c+a-b)x+(a+b-c)=0$
$\Delta =(c+a-b{)}^2-4(b+c-a\left)\right(a+b-c)$
$=(-2b{)}^2-4(-2a\left)\right(-2c)=4b^2-16ac=4(a-c{)}^2=4\left(\right(a+c{)}^2-4ac)=4(a-c{)}^2$
i.e $\Delta$ is a perfect square .
$\therefore$ Roots are rational.
$\therefore$ Roots are $\frac{-(c+a-b)\pm \sqrt{\Delta }}{2(b+c-a)}$ which are rational .
Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I .
Hence, option A .